PAT 1115 Counting Nodes in a BST

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A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than or equal to the node‘s key.
The right subtree of a node contains only nodes with keys greater than the node‘s key.
Both the left and right subtrees must also be binary search trees.
Insert a sequence of numbers into an initially empty binary search tree. Then you are supposed to count the total number of nodes in the lowest 2 levels of the resulting tree.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤1000) which is the size of the input sequence. Then given in the next line are the N integers in [?10001000] which are supposed to be inserted into an initially empty binary search tree.

Output Specification:
For each case, print in one line the numbers of nodes in the lowest 2 levels of the resulting tree in the format:

n1 + n2 = n

where n1 is the number of nodes in the lowest level, n2 is that of the level above, and n is the sum.

Sample Input:
9
25 30 42 16 20 20 35 -5 28

Sample Output:
2 + 4 = 6

#include<iostream> //建立二叉树和dfs
using namespace std;
int maxdeep=0, a=0, b=0;
struct node{
  int value;
  node* left=NULL;
  node* right=NULL;
  node(int v):value(v), left(NULL), right(NULL){
  }
};
node* insert(int v, node* root, int deep){
  if(!root){
    node* temp=new node(v);
    root =temp;
    maxdeep=(deep>maxdeep?deep:maxdeep);
  }else{
    if(v<=root->value)
        root->left=insert(v, root->left, deep+1);
    else
        root->right=insert(v, root->right, deep+1);
  }
  return root;
}
void preorder(node* root, int deep){    
    if(root==NULL)
        return;
    if(deep==maxdeep-1)
        a++;
    else if(deep==maxdeep)
        b++;
    preorder(root->left, deep+1);
    preorder(root->right, deep+1);
    
}
int main(){
  int n, v;
  cin>>n;
  node* root=NULL;
  for(int i=0; i<n; i++){
    cin>>v;
    root=insert(v, root, 1);
  }
  preorder(root, 1);
  cout<<b<<" + "<<a<<" = "<<a+b<<endl;
  return 0;
}

PAT 1115 Counting Nodes in a BST

标签：case   you   return   ted   tput   rop   ati   equal   space   

原文地址：https://www.cnblogs.com/A-Little-Nut/p/9502031.html